\subsection{Comparison test}
In this section, we will let \(a_n \in \mathbb R, a_n \geq 0\).
In other words, all series contain only non-negative real terms.
\begin{theorem}
	Suppose \(0 \leq b_n \leq a_n\) for all \(n\).
	If \(\sum_{j=1}^\infty a_j\) converges, then \(\sum_{j=1}^\infty b_j\) converges.
\end{theorem}
\begin{proof}
	Let \(s_N\) be the \(N\)th partial sum over the \(a_n\), and let \(d_N\) be the \(N\)th partial sum over the \(b_n\).
	Since \(b_n \leq a_n, d_N \leq s_N\).
	But \(s_N \to s\), so \(d_N \leq s_N \leq s\).
	So \(d_N\) is an increasing sequence that is bounded above by \(s\), so it converges.
\end{proof}
For example, let us analyse the behaviour of the sum of the sequence \(\frac{1}{n^2}\).
Note that
\[
	\frac{1}{n^2} < \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}
\]
for \(n \geq 2\).
By the comparison test, it is sufficient to show that the series on the right hand side converges, in order to show that the original series converges.
\[
	\sum_{j=2}^N a_j = 1 - \frac{1}{N} \to 1
\]
as required.
So the original series tends to some value less than or equal to 2.

\subsection{Cauchy's root test}
\begin{theorem}
	Suppose we have a sequence of non-negative terms \(a_n\).
	Suppose that \(a_n^{1/n} \to a\) as \(n \to \infty\).
	Then if \(a < 1\), the series \(\sum a_n\) converges.
	If \(a > 1\), the series \(\sum a_n\) diverges.
\end{theorem}
\begin{remark}
	Nothing can be said if \(a=1\).
	There is an example later of this fact.
\end{remark}
\begin{proof}
	If \(a < 1\), let us choose an \(r\) such that \(a < r < 1\).
	By the definition of the limit, \(\exists N\) such that \(\forall n \geq N\), \(a_n^{1/n} < r\).
	This implies that \(a_n < r^n\).
	The geometric series \(\sum r^n\) converges.
	By comparison, the series \(a_n\) converges.

	If \(a > 1\), for all \(n \geq N\), \(a_n^{1/n} > 1\) which implies \(a_n > 1\), thus \(\sum a_n\) diverges, since \(a_n\) does not tend to zero.
\end{proof}

\subsection{D'Alembert's ratio test}
\begin{theorem}
	Suppose \(a_n > 0\), and \(\frac{a_{n+1}}{a_n} \to \ell\).
	If \(\ell < 1\), then the series \(\sum a_n\) converges.
	If \(\ell > 1\), then the series \(\sum a_n\) diverges.
\end{theorem}
\begin{remark}
	Like before, no conclusion can be drawn if \(\ell = 1\).
\end{remark}
\begin{proof}
	Suppose \(\ell < 1\).
	We can choose \(\ell < r < 1\), \(\exists N\) such that \(\forall n \geq N\), \(\frac{a_{n+1}}{a_n} < r\).
	Therefore \(a_n < r^{n-N} a_N\).
	Hence, \(a_n < k r^n\) where \(k\) is independent of \(n\).
	Applying the comparison test, the series \(\sum a_n\) must converge.

	If \(\ell > 1\), we can choose \(\ell > r > 1\).
	Then \(\exists N\) such that \(\forall n \geq N\), \(\frac{a_{n+1}}{a_n} > r\).
	As before, \(a_n > r^{n-N} a_N\).
	But the \(r^{n-N}\) diverges, so the original series diverges.
\end{proof}

\begin{example}
Consider \(\sum_1^\infty \frac{n}{2^n}\).
We have
\[
	\frac{a_{n+1}}{a_n} = \frac{(n+1)/2^{n+1}}{n/2^n} \to \frac{1}{2}
\]
So we have convergence, by the ratio test.
Now, consider \(\sum_1^\infty \frac{1}{n}\) and \(\sum_1^\infty \frac{1}{n^2}\).
In both cases, the ratio test gives limit 1.
So the ratio test is inconclusive if the limit is 1.
Since \(n^{1/n} \to 1\), the root test is also inconclusive when the limit is 1.
To check this limit, we can write
\[
	n^{1/n} = 1 + \delta_n;\quad \delta_n > 0
\]
\[
	n = (1 + \delta_n)^n > \frac{n(n-1)}{2}\delta_n^2
\]
using the binomial expansion.
\[
	\implies \delta_n^2 < \frac{2}{n-1} \implies \delta_n \to 0
\]
The root test is a good candidate for series that contain powers of \(n\), for example
\[
	\sum_1^\infty \left[ \frac{n+1}{3n+5} \right]^n
\]
In this instance, for example, we have convergence.
\end{example}

\subsection{Cauchy's condensation test}
\begin{theorem}
	Let \(a_n\) be a decreasing sequence of positive terms.
	Then \(\sum_1^\infty a_n\) converges if and only if \(\sum_1^\infty 2^n a_{2^n}\) converges.
\end{theorem}
\begin{proof}
	First, note that if \(a_n\) is decreasing, then
	\[
		a_{2^k} \underset{(\ast)}{\leq} a_{2^{k-1} + i} \underset{(\dagger)}{\leq} a_{2^{k-1}};\quad 1 \leq i \leq 2^{k-1};\quad k \geq 1
	\]
	Now let us assume that \(\sum a_n\) converges to \(A \in \mathbb R\).
	Then, by \((\ast)\),
	\begin{align*}
		2^{n-1} a_{2^n} & = a_{2^n} + a_{2^n} + \dots + a_{2^n}                \\
		                & \leq a_{2^{n-1}+1} + a_{2^{n-1}+2} + \dots + a_{2^n} \\
		                & = \sum_{m=2^{n-1}+1}^{2^n}a_m
	\end{align*}
	Thus,
	\[
		\sum_{n=1}^N 2^{n-1}a_{2^n} \leq \sum_{n=1}^N \sum_{m=2^{n-1}+1}^{2^n} a_m = \sum_{n=2}^{2^N} a_m
	\]
	Therefore,
	\[
		\sum_{n=1}^N 2^n a_{2^n} \leq 2 \sum_{n=2}^{2^N} a_m \leq 2(A-a_1)
	\]
	Thus \(\sum_{n=1}^N 2^n a_{2^n}\) converges, since it is increasing and bounded above.
	For the converse, we will assume that \(\sum 2^n a_{2^n}\) converges to \(B\).
	Using \((\dagger)\),
	\begin{align*}
		\sum_{m=2^{n-1}}^{2^n} a_m & = a_{2^{n-1}} + a_{2^{n-1}+1} + \dots + a_{2^n}      \\
		                           & \leq a_{2^{n-1}} + a_{2^{n-1}} + \dots + a_{2^{n-1}} \\
		                           & = 2^{n-1}a_{2^{n-1}}
	\end{align*}
	So we have
	\[
		\sum_{m=2}^{2^N} a_m = \sum_{n=1}^N \sum_{m=2^{n-1}+1}^{2^n} a_m \leq \sum_{n=1}^N 2^{n-1} a_{2^{n-1}} \leq \frac{1}{2} B
	\]
	Therefore, \(\sum_{m=1}^N a_m\) is a bounded, increasing sequence and hence converges.
\end{proof}
Let us consider an example of this test.
Consider the series definition of the Riemann zeta function
\[
	\zeta(k) = \sum_{n=1}^\infty \frac{1}{n^k}
\]
For what \(k \in \mathbb R, k>0\) does this series converge?
This is equivalent to asking if the following series converges.
\[
	\sum_{n=1}^\infty 2^n \left[ \frac{1}{2^n} \right]^k = \sum_{n=1}^\infty \left( 2^{1-k} \right)^n
\]
Hence it converges if and only if \(2^{1-k} < 1 \iff k > 1\).

\subsection{Alternating series}
An alternating series is a series where the sign on each term switches between positive and negative.
\begin{theorem}[Alternating Series Test]
	If \(a_n\) decreases and tends to zero as \(u \to \infty\), then the alternating series
	\[
		\sum_1^\infty (-1)^{n+1} a_n
	\]
	converges.
\end{theorem}
\begin{proof}
	Let us consider the partial sum
	\[
		s_n = a_1 - a_2 + a_3 - a_4 + \dots + (-1)^{n+1}a_n
	\]
	In particular,
	\[
		s_{2n} = (a_1 - a_2) + (a_3 - a_4) + \dots + (a_{2n-1} - a_{2n})
	\]
	Since the sequence is decreasing, each parenthesised block is positive.
	Then \(s_{2n} \geq s_{2n-2}\).
	We can also write the partial sum as
	\[
		s_{2n} = a_1 - (a_2 - a_3) - (a_4 - a_5) - \dots - (a_{2n-2} - a_{2n-1}) - a_{2n}
	\]
	Each parenthesised block here is negative.
	So \(s_{2n} \leq a_1\).
	So \(s_{2n}\) is increasing and bounded above, so it must converge.
	Now, note that
	\[
		s_{2n+1} = s_{2n} + a_{2n+1} \to s_{2n}
	\]
	since \(a_{2n+1} \to 0\).
	So \(s_{2n+1}\) also converges, in fact to the same limit.
	Hence \(s_n\) converges to this same limit.
\end{proof}
